Cited passages: equations 10.7–10.20, decay law, half-life, mean lifetime, and activity.

The International System of Units base unit of measurement of time.

A unit of area used for scattering cross-sections.

A unit of time equal to $86400$ seconds.

A trypsin colorimetric assay follows enzyme activity through absorbance at $405\mathrm{nm}$. Trypsin cleaves an amide bond in the synthetic arginine derivative BAPNA, releasing p-Nitroaniline.^[ref]

The inspected Sigma procedure prepares fresh trypsin in cold $1\mathrm{mM}$ hydrochloric acid and fully dissolves $1mg/ml$ BAPNA. Its buffer solution stock contains $200\mathrm{mM}$ triethanolamine and $20\mathrm{mM}$ calcium chloride, at pH $7.8$. Three $3.00\mathrm{ml}$ cuvettes compare trypsin, trypsin plus the enzyme inhibitor aprotinin, and a blank replacing trypsin with its acid solvent. Sodium chloride solution replaces aprotinin in the uninhibited control. A thermostatted instrument for spectrophotometry holds $25{}^{\circ }\mathrm{C}$ and a $1\mathrm{cm}$ light path. After equilibration, adding BAPNA starts approximately five minutes of absorbance recording. The uninhibited sample checks activity; the blank checks background; the inhibited sample tests suppression.^[ref]

Let $s_U$, $s_I$ and $s_B$ denote the measured absorbance slopes per minute over the same linear time interval. Blank subtraction gives $s_U-s_B$ and $s_I-s_B$. If p-Nitroaniline accounts for the changing absorbance, the Beer-Lambert law givesHere $c$ is product molar concentration and $l$ is the geometric light path through the chemical solution. Sigma uses $\epsilon =9.96\mathrm{m}{\mathrm{M}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$.^[ref]

Choose original illustrative slopes $s_U=0.1036$, $s_I=0.0538$ and $s_B=0.0040\mathrm{mi}{\mathrm{n}}^{-1}$. The corrected uninhibited slope is $0.0996\mathrm{mi}{\mathrm{n}}^{-1}$, giving $dc/dt=0.0100\mathrm{mM}\mathrm{mi}{\mathrm{n}}^{-1}=10.0\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$ at $l=1\mathrm{cm}$. In $3.00\mathrm{ml}$, the product amount of substance increases at $0.0300\mu \mathrm{mol}\mathrm{mi}{\mathrm{n}}^{-1}$. The corrected inhibited slope is $0.0498\mathrm{mi}{\mathrm{n}}^{-1}$, half as large. The relative suppression is thereforeThe background cancels in the numerator, but remains essential in the denominator. Using uncorrected slopes would instead give about $48.07\%$. Differentiating $A_{405}=\epsilon lc$ with fixed $\epsilon$ and $l$ explains the rate conversion; dimensional analysis checks it independently, since $\mathrm{mi}{\mathrm{n}}^{-1}/(\mathrm{m}{\mathrm{M}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}\mathrm{cm})=\mathrm{mM}\mathrm{mi}{\mathrm{n}}^{-1}$.

The inference is a reaction rate on BAPNA under the stated conditions, rather than a direct count of trypsin molecules or proof of every cut in a protein. Cloudiness, other absorbing solutes, or different product protonation can compromise the Beer-Lambert law interpretation. A matched blank and a linear interval are therefore part of the measurement's meaning. A coloured enzyme inhibitor may require its own matched blank; a changed substrate or pH requires a fresh calibration. The constructed constant slopes above describe only that interval, not indefinite product accumulation.

Write a peptide's protein primary structure from its N-terminus toward its C-terminus. The usual simplified trypsin cleavage specificity rule places a cut immediately after K (lysine) or R (arginine), unless the next amino acid residue is P (proline). The following amino acid residue is essential: a terminal K or R has no following peptide bond to cut. The inspected institutional documentation describes proline as usually blocking cleavage, and explicitly lists sequence contexts in which that blocking is negligible. It also warns that experimental results can differ from sequence-based predictions. Thus “never before proline” is too strong.^[ref]

For an original worked example, stipulate complete cleavage at all sites allowed by that simplified rule in the twelve-amino acid residue sequenceA denotes alanine, G denotes glycine, and K, R and P have the meanings above. Cuts follow positions $3$, $7$ and $10$. Position $5$ is R, but its following P suppresses that cut in the stipulated rule; position $12$ is terminal K. Notice that the R in ARPK is retained inside the fragment, despite being a usually favoured trypsin target.

Near neutral pH, the lysine and arginine amino acid side chains undergo predominant protonation and carry positive electric charge. This helps distinguish those amino acid residues from the uncharged letter symbols used to write the sequence; those symbols specify identities, not whole-fragment electric charges.^[ref]

Fragment accounting checks the prediction: lengths $3+4+3+2=12$ conserve the twelve amino acid residues. The original linear chain has eleven peptide bonds; the four fragments retain $2+3+2+1=8$, so three were hydrolysed. Three water molecules are consumed per original chain in this formal stoichiometry. More generally, cutting $m$ distinct bonds of one linear peptide makes $m+1$ fragments: each cut splits one connected piece into two. This argument assumes an unbranched chain without additional links joining the pieces.

A missed trypsin cleavage at position $7$ would instead give $\mathrm{AGK}|\mathrm{ARPKAGR}|\mathrm{AK}$, merging two neighbouring fragments without changing their order. Even a correct cleavage preference does not guarantee complete proteolysis; predicted fragment boundaries describe a model, while actual boundaries require evidence from the digest.